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Pavel [41]
3 years ago
9

What is the mÐJAZ? plz help

Mathematics
1 answer:
Simora [160]3 years ago
8 0
X=3 Should be the answer If you ser 2x-3=5x-12
-2x -2x
-3=3x-12
+12 +12
9=3x
3=x
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Is y=3x-5 and 6x=2y+10 a solution
iVinArrow [24]
Y=3x-5
y-3x=-5
Times 2 from both side
y(2)-3x(2)=-5(2)
2y-6x=-10
Times negative from both side
(-)2y-6x(-)=(-)-10
-2y+6x=10
or
6x-2y=10


6x=2y+10
6x-2y=10
Furthermore, we see that both equation have the same 6x-2y=10 which means that both of them have infinity solutions, not a solution. Hope it help!
5 0
3 years ago
A kitchen drawer has a Volume 1,125 cubic inches. The drawer is 15 inches long and 5 inches deep. What is the width of the drawe
Temka [501]

Answer:

The width of the drawer is 15\ in

Step-by-step explanation:

we know that

The volume of the drawer is equal to

V=LWH

we have

L=15\ in

H=5\ in -----> the deep of the drawer

V=1,125\ in^{3}

substitute the given values in the formula and solve for W

1,125=(15)(W)(5)

1,125=(75)(W)

W=1,125/(75)

W=15\ in

7 0
3 years ago
7.3 as a mixed number simlified
joja [24]
Answer: 7 3/10
seven as a whole number, and .3 equals to 3/10 when converted
8 0
3 years ago
Use pascal's triangle to expand the following binomial expression <br>1.(2k-1/3)⁶​
Gwar [14]

9514 1404 393

Answer:

  64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

Step-by-step explanation:

The row of Pascal's triangle we need for a 6th power expansion is ...

  1, 6, 15, 20, 15, 6, 1

These are the coefficients of the products (a^(n-k))(b^k) in the expansion of (a+b)^n as k ranges from 0 to n.

Your expansion is ...

  1(2k)^6(-1/3)^0 +6(2k)^5(-1/3)^1 +15(2k)^4(-1/3)^2 +20(2k)^3(-1/3)^3 +...

     15(2k)^2(-1/3)^4 +6(2k)^1(-1/3)^5 +1(2k)^0(-1/3)^6

  = 64k^6 -64k^5 +(80/3)k^4 -(160/27)k^3 +(20/27)k^2 -(4/81)k +1/729

7 0
2 years ago
Could use some help please
Helen [10]

Answer:

Area of Square= 12*12= 144

Area of Circle= πr2

Radius= 12/2= 6

Area of Circle= 3.14*6*6 =113.04

Area of shaded region= 144- 113.04 = 30.96 = 31.0 (1 dp)

5 0
2 years ago
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